Question Number 100557 by Mikael_786 last updated on 27/Jun/20

Ω=∫_0 ^∞ (e^(ax) /(e^(bx) +1))dx, b>a

Answered by mathmax by abdo last updated on 27/Jun/20

Ω =∫_0 ^∞ (e^(ax) /(e^(bx) +1))dx⇒Ω =∫_0 ^∞ (e^((a−b)x) /(1+e^(−bx) ))dx =∫_0 ^∞ e^(−(b−a)x) Σ_(n=0) ^∞ (−1)^n e^(−nbx) dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ e^(−(b−a+nb)x) dx =Σ_(n=0) ^∞ (−1)^n [((−1)/(b−a+nb)) e^(−(b−a+nb)x) ]_0 ^∞ =Σ_(n=0) ^∞ (((−1)^n )/(b−a +nb)) ....be continued....

Answered by mathmax by abdo last updated on 27/Jun/20

let try another way we do the changement e^(bx) =t ⇒bx =lnt ⇒x =((lnt)/b) ⇒ Ω =∫_0 ^∞ (e^((a/b)lnt) /(t+1)) ×(dt/(bt)) =(1/b)∫_0 ^∞ (t^((a/b)−1) /(t+1))dt for 0<(a/b)<1 we get Ω =(1/b)×(π/(sin(((πa)/b)))) ( i have used ∫_0 ^∞ (t^(α−1) /(1+t))dt =(π/(sin(πα)))) ⇒ Ω =(π/(b sin(((πa)/b))))

Commented byMikael_786 last updated on 28/Jun/20

thank you